Peter4075
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 Sep 21 reviewed Approve Getting started self-studying general relativity Sep 4 comment Can a pilot in a small spaceship feel G force in space? @CuriousOne - Honing "physics intuition" needn't involve flip, superior comments. Not the best way to encourage learners to use this site. Sep 4 comment Can a pilot in a small spaceship feel G force in space? @CuriousOne - that is not really a helpful comment. Aug 29 accepted Derivation of geodesic deviation equation from two neighbouring geodesics Aug 29 comment Derivation of geodesic deviation equation from two neighbouring geodesics Thanks. I would never have seen that. Aug 28 comment Derivation of geodesic deviation equation from two neighbouring geodesics Thanks, but I still don't see why something multiplied by the derivative of that something should be second order. How do we know that $\frac{d\xi^{c}}{du}$ is a small number? I'm assuming second order in this derivation means something squared, not a second order differential equation. Aug 28 asked Derivation of geodesic deviation equation from two neighbouring geodesics Jul 19 awarded Disciplined Jun 25 awarded Popular Question Jun 24 awarded Enthusiast Jun 17 awarded Yearling Jun 16 comment Nabla or semicolon notation for covariant derivative? I guess saying it's just a matter of opinion is an answer of sorts. Jun 16 asked Nabla or semicolon notation for covariant derivative? Jun 6 awarded Nice Question Jun 2 comment How are these two Riemann tensor equations equivalent? Yes, I see it now. Thanks. Jun 2 accepted How are these two Riemann tensor equations equivalent? Jun 2 comment How are these two Riemann tensor equations equivalent? OK, I multiply by $g^{ax}$ (?) to give$$\lambda_{;bc}^{x}-\lambda_{;cb}^{x}=R_{\phantom{\mu}bc}^{dx}\lambda_{d}.$$ Then what? Jun 2 comment How are these two Riemann tensor equations equivalent? OK, so what metric do I multiply by? If I try $g^{ax}$ it does weird things to the rhs (ie I have two upper indices in the Riemann tensor). And I'm still left with $\lambda^{a}$. Jun 2 comment How are these two Riemann tensor equations equivalent? So I multiply by $g^{ad}$ to give$$\lambda_{;bc}^{d}-\lambda_{;cb}^{d}=R_{\phantom{\mu}abc}^{d}\lambda^{a}.$$ Then use the Riemann tensor symmetries to give$$\lambda_{;bc}^{d}-\lambda_{;cb}^{d}= -R_{\phantom{\mu}dbc}^{a}\lambda^{a}.$$ But it's still not the same as Poisson's equation. Jun 2 revised How are these two Riemann tensor equations equivalent? edited tags